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07.rb
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# --- Day 7: The Sum of Its Parts ---
#
# You find yourself standing on a snow-covered coastline; apparently,
# you landed a little off course. The region is too hilly to see the
# North Pole from here, but you do spot some Elves that seem to be
# trying to unpack something that washed ashore. It's quite cold out,
# so you decide to risk creating a paradox by asking them for
# directions.
#
# "Oh, are you the search party?" Somehow, you can understand
# whatever Elves from the year 1018 speak; you assume it's Ancient
# Nordic Elvish. Could the device on your wrist also be a translator?
# "Those clothes don't look very warm; take this." They hand you a
# heavy coat.
#
# "We do need to find our way back to the North Pole, but we have
# higher priorities at the moment. You see, believe it or not, this
# box contains something that will solve all of Santa's transportation
# problems - at least, that's what it looks like from the pictures in
# the instructions." It doesn't seem like they can read whatever
# language it's in, but you can: "Sleigh kit. Some assembly required."
#
# "'Sleigh'? What a wonderful name! You must help us assemble this
# 'sleigh' at once!" They start excitedly pulling more parts out of
# the box.
#
# The instructions specify a series of steps and requirements about
# which steps must be finished before others can begin (your puzzle
# input). Each step is designated by a single letter. For example,
# suppose you have the following instructions:
#
# Step C must be finished before step A can begin.
# Step C must be finished before step F can begin.
# Step A must be finished before step B can begin.
# Step A must be finished before step D can begin.
# Step B must be finished before step E can begin.
# Step D must be finished before step E can begin.
# Step F must be finished before step E can begin.
#
# Visually, these requirements look like this:
#
# -->A--->B--
# / \ \
# C -->D----->E
# \ /
# ---->F-----
#
# Your first goal is to determine the order in which the steps should
# be completed. If more than one step is ready, choose the step which
# is first alphabetically. In this example, the steps would be
# completed as follows:
#
# - Only C is available, and so it is done first.
# - Next, both A and F are available. A is first alphabetically, so
# it is done next.
# - Then, even though F was available earlier, steps B and D are now
# also available, and B is the first alphabetically of the three.
# - After that, only D and F are available. E is not available
# because only some of its prerequisites are complete. Therefore, D
# is completed next.
# - F is the only choice, so it is done next.
# - Finally, E is completed.
#
# So, in this example, the correct order is CABDFE.
#
# In what order should the steps in your instructions be completed?
require 'set'
class Step
attr_reader :id, :parents
@@all = {} # id => Step
def Step.lookup(id)
@@all.fetch(id) { new(id) }
end
@@ready = SortedSet.new # steps that are ready to be worked on (and
# also, in part 2, steps being worked on)
def Step.ready
@@ready
end
def initialize(id)
@id = id
@children = []
@parents = []
@completed = false
@@all[id] = self
@@ready.add(self)
end
def has_child(child)
@children.push(child)
child.has_parent(self)
end
def has_parent(parent)
@parents.push(parent)
@@ready.delete(self)
end
def completed?
@completed
end
def complete
@completed = true
@@ready.delete(self)
@children.each do |c|
@@ready.add(c) if c.parents.all? {|p| p.completed? }
end
end
def <=>(other)
@id <=> other.id
end
end
open("07.in").each do |l|
m = /^Step (.) must be finished before step (.) can begin\.$/.match(l)
Step.lookup(m[1]).has_child(Step.lookup(m[2]))
end
while Step.ready.length > 0
s = Step.ready.first
s.complete
print s.id
end
puts
# --- Part Two ---
#
# As you're about to begin construction, four of the Elves offer to
# help. "The sun will set soon; it'll go faster if we work together."
# Now, you need to account for multiple people working on steps
# simultaneously. If multiple steps are available, workers should
# still begin them in alphabetical order.
#
# Each step takes 60 seconds plus an amount corresponding to its
# letter: A=1, B=2, C=3, and so on. So, step A takes 60+1=61 seconds,
# while step Z takes 60+26=86 seconds. No time is required between
# steps.
#
# To simplify things for the example, however, suppose you only have
# help from one Elf (a total of two workers) and that each step takes
# 60 fewer seconds (so that step A takes 1 second and step Z takes 26
# seconds). Then, using the same instructions as above, this is how
# each second would be spent:
#
# Second Worker 1 Worker 2 Done
# 0 C .
# 1 C .
# 2 C .
# 3 A F C
# 4 B F CA
# 5 B F CA
# 6 D F CAB
# 7 D F CAB
# 8 D F CAB
# 9 D . CABF
# 10 E . CABFD
# 11 E . CABFD
# 12 E . CABFD
# 13 E . CABFD
# 14 E . CABFD
# 15 . . CABFDE
#
# Each row represents one second of time. The Second column
# identifies how many seconds have passed as of the beginning of that
# second. Each worker column shows the step that worker is currently
# doing (or . if they are idle). The Done column shows completed
# steps.
#
# Note that the order of the steps has changed; this is because steps
# now take time to finish and multiple workers can begin multiple
# steps simultaneously.
#
# In this example, it would take 15 seconds for two workers to complete
# these steps.
#
# With 5 workers and the 60+ second step durations described above,
# how long will it take to complete all of the steps?
class Step
def reset
@completed = false
@@ready.add(self) if @parents.empty?
end
def Step.reset
@@all.values.each do |s|
s.reset
end
end
def start
@counter = id.ord-"A".ord + 61
end
def started?
defined?(@counter)
end
def advance
@counter -= 1
complete if @counter == 0
end
end
class Worker
def initialize
@step = nil
end
def assign(step)
@step = step
step.start
end
def idle?
@step.nil?
end
def working?
!idle?
end
def work
@step.advance
@step = nil if @step.completed?
end
end
Step.reset
workers = Array.new(5) { Worker.new }
time = 0
while Step.ready.length > 0
workers.select {|w| w.idle? }.each do |w|
s = Step.ready.select {|s| !s.started? }.min {|s,t| s.id <=> t.id }
w.assign(s) if !s.nil?
end
workers.each {|w| w.work if w.working? }
time += 1
end
puts time